Smallest distance between object and image - lens.

In summary, the distance between the object and a real image is the smallest when the focal distance of the lens is f.
  • #1
Kruum
220
0

Homework Statement



Find the smallest distance between the object and a real image, when the focal distance of the lens is [tex]f[/tex]

Homework Equations



[tex] \frac{1}{s}+ \frac{1}{p} = \frac{1}{f}[/tex], where [tex]s[/tex] is the distance of the object from the lens and [tex]p[/tex] is the distance of the image.

The Attempt at a Solution



I'm not even sure, what I'm trying to do here, since the definition of a real and virtual image is a bit vague to me. But I've got something...

Let [tex]D=s+p[/tex] be the desired distance. From the equation above we get [tex]p = \frac{sf}{s-f}[/tex] so [tex]D= \frac{s^2}{s-f}[/tex]. Then [tex] \frac{ \partial D}{ \partial s}= \frac{2s^2-2sf-s^2}{(s-f)^2}=0 \Rightarrow s=0[/tex] or [tex]s=2f[/tex]. We get the same for [tex]p[/tex], so the distance would be [tex]D=4f[/tex]. Is this correct and does this apply for the minimum distance only or are the real image and the object always equal length from the lens?
 
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  • #2
Kruum said:

The Attempt at a Solution



I'm not even sure, what I'm trying to do here, since the definition of a real and virtual image is a bit vague to me. But I've got something...

Let [tex]D=s+p[/tex] be the desired distance. From the equation above we get [tex]p = \frac{sf}{s-f}[/tex] so [tex]D= \frac{s^2}{s-f}[/tex]. Then [tex] \frac{ \partial D}{ \partial s}= \frac{2s^2-2sf-s^2}{(s-f)^2}=0 \Rightarrow s=0[/tex] or [tex]s=2f[/tex]. We get the same for [tex]p[/tex], so the distance would be [tex]D=4f[/tex]. Is this correct and does this apply for the minimum distance only or are the real image and the object always equal length from the lens?

You have just arrived at this value as a condition for an extremum (or a stationary value). What do you feel should be the answer? The value is correct, by the way . Can you find values of s and p such that [itex]s+p\neq4f[/itex]?

Not understanding the distinction between a real and a virtual image is not very conducive to learning Physics. Please read up or ask for help here.
 
  • #3
Shooting Star said:
Not understanding the distinction between a real and a virtual image is not very conducive to learning Physics. Please read up or ask for help here.

That's the thing, I have read and listened and either misunderstood or then have had contradicting information. First I learned that the magnification for real image is m<0, the image is up-side down, and for virtual image m>0. Now I'm taught that virtual image always appears on the other side of the lens (or mirror). For converging lenses the image appears on the other side of the lens with m<0. :uhh:
 
  • #5
hi there! I'm someone else asking the same question. so is the answer 4f? I don't understand what you mean by extremum and find the values of s and p so that it cannot equal 4f.
 

1. What is the smallest distance between an object and its image formed by a lens?

The smallest distance between an object and its image formed by a lens is known as the focal length. It is the distance between the lens and the point where the light rays converge to form a sharp image.

2. How is the smallest distance between object and image affected by the properties of the lens?

The smallest distance between object and image is affected by the properties of the lens, such as its focal length and aperture size. A lens with a shorter focal length will have a smaller distance between the object and its image, while a larger aperture will allow for a shorter distance.

3. Can the smallest distance between object and image be changed?

Yes, the smallest distance between object and image can be changed by adjusting the properties of the lens, such as the focal length and aperture size. It can also be changed by using additional lenses or optical devices.

4. How does the distance between the object and lens affect the image?

The distance between the object and lens can affect the size and clarity of the image. If the object is placed too close to the lens, it may appear blurry or distorted. However, if the object is placed at the correct distance, the image will be sharp and clear.

5. What is the relationship between the smallest distance between object and image and magnification?

The smallest distance between object and image is directly related to the magnification of the image. A shorter distance will result in a larger magnification, while a longer distance will result in a smaller magnification. This is because the closer the object is to the lens, the larger the resulting image will be.

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